Question: Divide the following complex numbers. $\dfrac{-8+24i}{4-4i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${4+4i}$. $ \dfrac{-8+24i}{4-4i} = \dfrac{-8+24i}{4-4i} \cdot \dfrac{{4+4i}}{{4+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-8+24i) \cdot (4+4i)} {4^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-8+24i) \cdot (4+4i)} {(4)^2 - (-4i)^2} $ $ = \dfrac{(-8+24i) \cdot (4+4i)} {16 + 16} $ $ = \dfrac{(-8+24i) \cdot (4+4i)} {32} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8+24i}) \cdot ({4+4i})} {32} $ $ = \dfrac{{-8} \cdot {4} + {24} \cdot {4 i} + {-8} \cdot {4 i} + {24} \cdot {4 i^2}} {32} $ $ = \dfrac{-32 + 96i - 32i + 96 i^2} {32} $ Finally, simplify the fraction. $ \dfrac{-32 + 96i - 32i - 96} {32} = \dfrac{-128 + 64i} {32} = -4+2i $